A galvanometer having a coil of $12\Omega $ gives full scale deflection for a current of $4$mA. How can it be converted into a voltmeter of range 0 to 24 V?

Hint: A galvanometer is an electromagnetic device that can be used to measure currents even up to the order of a few microamperes. This galvanometer device is used to detect the presence of current or measure its magnitude in a circuit. To solve this question use the equation of voltage drop across a branch consisting of a series combination of the galvanometer and a high resistance. Then by substituting the value of galvanometer resistance, the current through the coil and the voltmeter value we can determine the value of the resistance.

Complete step-by-step solution:
The given values are:
Resistance of galvanometer $R_G$ $ = 12\Omega $
Current of galvanometer I$_G$=$4$mA= $4*{10^{ - 3}}$A
Voltage= $24$ V
To find: Resistance R
Using the equation: \[V = {\text{ }}{I_G}\left( {R + {R_G}} \right)\]
Rearranging we get, \[\dfrac{V}{{{I_G}}} = \left( {R + {R_G}} \right)\]
$R = \dfrac{V}{{{I_G}}} - {R_G}$
Substituting the values of resistance, voltage and current we get, $R = \dfrac{{24}}{{4*{{10}^{ - 3}}}} - 12$
Simplifying the values we get, $R = \dfrac{6}{{{{10}^{ - 3}}}} - 12 = 6000 - 12 = 5988\Omega $
To be converted into a voltmeter of range $0 - 24$ V there needs to be a resistance of $5988\Omega$

Note: While doing the calculations care must be taken to convert the unit of current from the given milliAmperes to Amperes. This conversion can be done by multiplying the milli Ampere value by ${10^{ - 3}}$, this will give the final value in Amperes. If the values are not converted to the SI units or if all the values are not in the same unit then our calculations can go wrong. The advantage of a galvanometer is that they have high sensitivity, high accuracy and it is not affected by stray magnetic fields. However, it can be used only for direct currents and result in errors due to damage of spring from mechanical stress.